A calorimeter of water equivalent 20\; g contains 180 \; g of water at 25^{\circ}C. 'm' grams of steam at 100^{\circ}C is mixed in it till the temperature of the mixture is 31^{\circ}C. The value of 'm' is close to (Latent heat of water =540\; cal\; g^{-1}, specific heat of water =1\; cal\; g^{-1}\; ^{\circ}C^{-1} )
Option: 1 2
Option: 2 4
Option: 3 3.2
Option: 4 2.6

Answers (1)

As we know

Q=mc\Delta T

and as 

Heat lost by steam = heat gained by the water

\begin{array}{l} [180 \times 1 \times(31-25)]+[20 \times(31-25)]=[m \times 540]+[m \times 1 \times(100-31)] \\ \Rightarrow (180 \times 6)+(20 \times 6)=540 m+100 m-31 m \\ \Rightarrow 1080+120=640 m-31 m \\ \Rightarrow 1200=609 m \\ \Rightarrow m=\frac{1200}{609}=1.97\approx 2 \end{array}

 

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